Question

The eighth term of an arithmetic sequence is 40. Calculate the sum of 15 terms

Answer 1

a8 = 40

a+ (8-1)d = 40

a+7d = 40..........(1)

now, we have to find S15

n = 15

Sn = n/2[a+an]

Sn = n/2[a+a+(n-1)d]

S15 = 15/2[2a+(15-1)d]

S15 = 15/2[2a+14d]

S15=( 15/2)×2[a+7d]

S15= 15[a+7d]

S15 = 15×40.........from (1)

S15 = 600

Answer 2

Answer:

The sum of 15 terms of an AP whose eighth term is 40 is 600.

Step-by-step explanation:

As per the question,

We need to find the sum of 15 terms of an AP whose eighth term is 40.

As we know, GIVEN THAT:

8th term of AP = 40

Applying fomula:

$a+(n-1)d=Tn\\a+(8-1)d=40\\a+7d=40$..............(i)

Now, to calculate the sum of 15 terms:

$Sn = \frac{n}{2} (T1 + Tn)\\\\Sn=\frac{15}{2} (a+a+(n-1)d)\\\\Sn=\frac{15}{2} (a+a+(15-1)d)\\\\\\Sn=\frac{15}{2} (2a+14d)\\\\\\Sn= 15(a+7d)\\Sn=15\times40=600$

Hence,

The sum of 15 terms of an AP whose eighth term is 40 is 600.

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